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2. Add Two Numbers - Medium

You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Example:

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

 

 

用一個dummy node記錄相加後的結果,同時還需要一個node記錄dummy linked list的頭,用以最後return

在迴圈裡,用sum和remainder分別表示兩位的加和、進位值。注意最後如果進位值不為0,還需要增加一個新的listnode

注意dummy linked list的處理:dummy.next = new listnode(xxx), dummy = dummy.next

time: O(n), space: O(n)

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        int remainder = 0;
        ListNode dummy 
= new ListNode(0); ListNode prehead = dummy; while(l1 != null || l2 != null) { int sum = remainder; if(l1 != null) { sum += l1.val; l1 = l1.next; } if(l2 != null) { sum += l2.val; l2 = l2.next; } ListNode tmp = new ListNode(sum % 10); remainder = sum / 10; dummy.next = tmp; dummy = dummy.next; } if(remainder != 0) { dummy.next = new ListNode(remainder); } return prehead.next; } }